Partial Differential equations Robin Eigenvalue problem If I have a Robin eigenvalue problem:
$X'-a_0X=0$ at $x=0$ and
$X' + a_lX=l$ at $x=l$ where $a_0$ and $a_l$ are given constants.  I'm assuming that $a_0<0$, $a_l<0$ and $-a_0-a_l<a_0a_ll$  I'm supposed to:
a) Show that there are two negative eigenvalues.
b) Show there are an infinite number of positive eigenvalues?
c) State whether or not there are zero or complex eigenvalues and
d) Use a,b,c to solve the diffusion equation with Robin boundary conditions:
$u_t=ku_{xx}$ for $0<x<l$, $0<t<\infty$
$u_x - a_0u=0$ for $x=0$, $u_x + a_lu=0$ for $x=l$
$u(x,0)=\phi(x)$
My solutions:
a) I was able to show there is one negative eigenvalue, and the missing positive eigenvalue appears as a negative eigenvalue and therefore there are 2 negative eigenvalues
b) I used a graph to show there are infinitely many positive eigenvalues
c) I'm struggling answer c, any suggestions?
d) This is what I have for d) but I doesn't seem correct to me:
Since there is two negative eigenvalues, -$\gamma_0^2$ and -$\gamma_1^2$ and a series of positive eigenvalues $\lambda_n=\beta_n^2>0$ for $n=1,2,3,...$  
So the solution is:
$u(x,t)=A_0e^{+\gamma_0^2kt}(cos(\gamma_0x + \frac{a_0}{\gamma_0}sinh(\gamma_0x)$ + $\sum\limits_{n=1}^{\infty}A_ne^{-\beta_n^2kt}(cos(\beta_nx) + \frac{a_0}{\beta_n}sin(\beta_nx)$
I'm trying to figure out how I find c) and if I did d) correctly?  Thanks for any and all help!
 A: c) The case of zero eigenvalue. Assume that $\gamma = 0$ is an eigenvalue of your problem. Hence, there must be a solution of the problem
$$
\left\{
\begin{aligned}
X'' &= \gamma X = 0,\\
X'(0) &- a_0 X(0) = 0,\\
X'(l) &+ a_l X(l) = l.
\end{aligned}
\right.
$$
Obviously, the general solution of $X'' = 0$ is 
$$
X = C_1 x + C_2.
$$
The question now is: are there $C_1$ and $C_2$, such that $C_1 x + C_2$ satisfies the given Robin boundary conditions? Lets try to find them. Note that $X' = C_1$. Hence,
$$
\left\{
\begin{aligned}
C_1 - a_0 C_2 = 0, \\
C_1 + a_l (C_1 l + C_2) = l.
\end{aligned}
\right.
\quad \Longrightarrow \quad 
\left\{
\begin{aligned}
C_1 &= a_0 C_2, \\
C_2 &= \frac{l}{a_0 + a_l + a_0 a_l l}.
\end{aligned}
\right.
$$
From your assumption $-a_0 - a_l < a_0 a_l l$ it follows that $C_1$ and $C_2$ exist and are defined uniquely. Hence, $X = C_1 x + C_2$ is an eigenfunction, which corresponds to the zero eigenvalue.
P.S. I don't understand what is mean by complex eigenvalue here. Is it the case $\gamma \in \mathbb{C}$? Or complex eigenvalues in the sense of corresponding matrix?
